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Estimate Instantaneous Velocity

  1. Jan 11, 2017 #1
    1. The problem statement, all variables and given/known data
    If a ball is thrown into the air with a velocity of 48 ft/s, its height in feet t seconds later is given by y = 48t − 16t2.
    (a) Find the average velocity for the time period beginning when t = 2 and lasting for each of the following.

    (i) 0.5 seconds

    (ii) 0.1 seconds

    (iii) 0.05 seconds

    (iv) 0.01 seconds

    (b) Estimate the instantaneous velocity when t = 2.

    2. Relevant equations

    [itex]Average Velocity =\dfrac{\triangle y}{\triangle t}[/itex] (Change in position over change in time)

    [itex]m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}[/itex]

    3. The attempt at a solution

    Set: [itex]t = 2[/itex] and [itex]t = 2 + h[/itex] (let h be the time difference) [itex]\ne 0[/itex]

    [itex]t = 2[/itex] is [itex]y_{1},x_{1}[/itex] and [itex]t = 2 + h[/itex] is [itex]y_{2},x_{2}[/itex]

    Average Velocity = [itex]\dfrac{y(2+h)-y(2)}{(2+h)-2}[/itex]

    Now with [itex]y=48t-16t^2[/itex], we will replace [itex]t[/itex] with [itex]2[/itex] and [itex]2 + h[/itex].

    [itex]\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2)]}{h}[/itex]

    [itex]\dfrac{96+48h-16(4+4h+h^2)-96+64}{h}[/itex]

    [itex]\dfrac{48h-64-64h-16h^2+64}{h}[/itex]

    [itex]\dfrac{-16h^2-16h}{h}[/itex]

    [itex]\dfrac{h(-16h-16)}{h}[/itex]

    Average Velocity =[itex]-16h-16[/itex]

    Then I plug in (i) 0.5 into the equation like this: [itex]-16(0.5)-16=24[/itex]
    I do this for the rest of them and get correct answers according to the online "WebAssign" website. Now I'm stuck on (b) "Estimate the instantaneous velocity when t = 2." You can look at my picture, I tried plugging in values near 2 like 2.1 and 2.01, etc, into [itex]y=48t-16t^2[/itex] What am I doing wrong? So as I get closer to 2 I get closer to the slope of 31.999 which then you make an educated guess (by rounding) and get 32. But this is wrong. I don't know why? Also, I heard there is another method for finding average velocity, my book doesn't have it--anyone know it? Much appreciation.
     

    Attached Files:

  2. jcsd
  3. Jan 11, 2017 #2

    Mark44

    Staff: Mentor

    For the instantaneous velocity near t = 2, look at the change in position divided by the change in time.

    Look at ##\frac{y(2 + h) - y(2)}{2 + h - 2}##, for some values of h that are close to 0, say h = .1, h = .01, h = .001.
    You should get a value close to -16 ft/sec.
     
  4. Jan 12, 2017 #3
    My teacher also said use values like 0.1, 0.01, etc. Why? I don't get why we're using those values? Also, how do we make the equation where we can plug in those values? from using [itex]\dfrac{y(2+h)-y(2)}{(2+h)-(2)}[/itex]
     
    Last edited: Jan 12, 2017
  5. Jan 12, 2017 #4

    Mark44

    Staff: Mentor

    Because you want the time interval to be very short so as to get a closer estimate of the instantaneous velocity.
    You have the equation: ##y(t) = 48t - 16t^2##
     
  6. Jan 12, 2017 #5
    This is where I don't understand. It says find instantaneous velocity when t = 2. Okay, so don't we want to use values close to 2? Like 2.1, 2.01. So if it was like, "find instantaneous velocity closer to when t = 5", would I still use 0.1, 0.01, etc? Because that means we are 0.1 from 5 or 0.01 from 5?

    When I plug [itex]48(t)-16(t)^2[/itex] into [itex]\dfrac{y(2+h)-y(2)}{(2+h)-2}[/itex], I don't understand how to do it. Am I actually plugging in [itex]48(t)-16(t)^2[/itex] into the y values of the other equation? Something like this:

    [itex]\dfrac{48t-16t^2(2+h)-48t-16t^2(2)}{2+h-2}[/itex]? Then distribute and simplify it down more?
     
  7. Jan 12, 2017 #6

    Mark44

    Staff: Mentor

    Yes. If h = .01, then 2 + h = 2 + .01 = 2.01
    No, you are "plugging in" 2 + h and 2, respectively, into your height function y(t) = 48t - 16t2
    For y(2 + h), replace t in your formula with 2 + h.
    For y(2), replace t with 2.
    Then simplify the result.
     
  8. Jan 12, 2017 #7
    [itex]48(2+h)-16(2)^2[/itex]

    [itex]96+48h-64[/itex]

    [itex]48h-32[/itex]

    Am I doing this right?
     
  9. Jan 12, 2017 #8

    Mark44

    Staff: Mentor

    I don't know what you're doing above.
    No. This is what you should be evaluating: ##\frac{y(2+h)-y(2)}{(2+h)-2}##, using ##y(t) = 48t - 16t^2##. The quotient here gives you the average speed between time t = 2 + h and time t = 2.
    Edit: Fixed a typo I had in the y(t) formula.
     
    Last edited: Jan 12, 2017
  10. Jan 12, 2017 #9
    How did you get [itex]y(t)=48t-32t^2[/itex]?

    Also, in above when you said "For y(2 + h), replace t in your formula with 2 + h. For y(2), replace t with 2." Isn't that what I did and got the [itex]48h - 32[/itex]?
     
  11. Jan 12, 2017 #10

    Mark44

    Staff: Mentor

    I miswrote 32t2 instead of 16t2. I have fixed my typo.

    Whatever you did isn't right. y(2 + h) gives you two terms, and y(2) also gives you two terms. Also, you have 2 + h - 2 in the denominator.
     
  12. Jan 12, 2017 #11

    Mark44

    Staff: Mentor

    Let's back up a few steps. You need to evaluate ##\frac{y(2 + h) - y(2)}{2 + h - 2}##, using ##y(t) = 48t - 16t^2##.

    What do you get for y(2 + h)?
    What do you get for y(2)?
     
  13. Jan 12, 2017 #12
    Am I placing [itex]48t-16t^2[/itex] into the y part of both [itex]y(2+h)[/itex] and [itex]y(2)[/itex]?

    Like this maybe:

    [itex]\dfrac{48(2+h)-16(2+h)^2-[48(2)-16(2)^2]}{(2+h)-2}[/itex]

    [itex]\dfrac{96+48h-16(h^2+4h+4)-96+64}{(2+h)-2}[/itex]

    [itex]\dfrac{96+48h-16h^2-64h-64-96+64}{(2+h)-2}[/itex]

    [itex]\dfrac{-16h^2-16h}{(2+h)-2}[/itex]

    I'm pretty sure something isn't right but I gave it a go here.
     
    Last edited: Jan 12, 2017
  14. Jan 12, 2017 #13

    Mark44

    Staff: Mentor

    This is correct, but you should simplify the denominator.

    When you've done that, estimate the velocity at a few times close to 2 sec., such as 2.1 (h = .1), 2.01 (h = .01), and however more you need to do.
     
  15. Jan 12, 2017 #14
    Will it just be [itex]\dfrac{-16h^2-16h}{h}[/itex]?
     
  16. Jan 12, 2017 #15

    Mark44

    Staff: Mentor

    And simplfy that...
     
  17. Jan 12, 2017 #16
    Like this:

    [itex]\dfrac{-16h(h+1)}{h}[/itex] Factor

    [itex]16(h+1)[/itex] Cancel out the h
     
  18. Jan 12, 2017 #17

    Mark44

    Staff: Mentor

    Where did the minus sign go?

    When you get that straigtened out, pick a few (small) values of h, as described in parts i) through iv) of the first post, and then estimate the velocity at t = 2 seconds.
     
  19. Jan 12, 2017 #18
    Oh, I forgot about that. It should've been there as [itex]-16(h+1)[/itex]

    i.)0.5

    [itex]-16(0.5+1) = -24 ft/s[/itex]​

    ii.)0.1

    [itex]-16(0.1+1) = -17.6 ft/s[/itex]​

    iii.)0.05

    [itex]-16(0.05+1) = -16.8 ft/s[/itex]​

    iv.)0.01

    [itex]-16(0.01+1) = 1-6.16 ft/s[/itex]​

    b.) Estimate the instantaneous velocity when t = 2.

    [itex]-16(0.1+1) = -17.6 ft/s[/itex]

    [itex]-16(0.01+1) = -16.16 ft/s[/itex]

    [itex]-16(0.001+1) = -16.08 ft/s[/itex]

    Estimate: [itex]-16 ft/s[/itex].

    I entered that in and got it right! Thank you very much for your help!
     
  20. Jan 12, 2017 #19

    Mark44

    Staff: Mentor

    Surely, you mean -16.16 ft/sec
     
  21. Jan 12, 2017 #20
    Oh, oops. Yeah, -16.16ft/s.
     
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